Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec.

Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec. Correct Answer 1.67

N flux of A = D(AB)/z * (Concentration difference) Concentration difference = 5/3 =1.67.
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